Question

A dice is thrown 2n+1 times, nϵn. The probability that faces with even numbers show odd number of times is

Answer 1

For a dice thrown 3(n=1) times. The probability of getting even numbers, an odd number of times, will have either one even number or 3 even numbers.

[math]Probability = (P(even) * P(not even) * P(not even)) + (P(even) * P(even) * P(even))[/math]

[math]So, P(X=1) =( \frac{1}{2} * \frac{1}{2} * \frac{1}{2} ) + ( \frac{1}{2} * \frac{1}{2} * \frac{1}{2} ) = \frac{2}{2^3}[/math]

Similarly for [math]n = 2[/math],

[math]P(X=2)= \frac{3}{2^5}[/math]

[math]P(X=3) = \frac{4}{2^7}[/math]

[math]P(X=n)= \frac{n+1}{2^{2n+1}}.